package editor.cn;

//Given an array nums. We define a running sum of an array as runningSum[i] = su
//m(nums[0]…nums[i]). 
//
// Return the running sum of nums. 
//
// 
// Example 1: 
//
// 
//Input: nums = [1,2,3,4]
//Output: [1,3,6,10]
//Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4]. 
//
// Example 2: 
//
// 
//Input: nums = [1,1,1,1,1]
//Output: [1,2,3,4,5]
//Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+
//1+1+1]. 
//
// Example 3: 
//
// 
//Input: nums = [3,1,2,10,1]
//Output: [3,4,6,16,17]
// 
//
// 
// Constraints: 
//
// 
// 1 <= nums.length <= 1000 
// -10^6 <= nums[i] <= 10^6 
// Related Topics 数组 
// 👍 41 👎 0

public class RunningSumOf1dArray{
    public static void main(String[] args) {
        Solution solution = new RunningSumOf1dArray().new Solution();
        
    }

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public int[] runningSum(int[] nums) {
        int[] results = new int[nums.length];
        for (int i = 0; i <nums.length ; i++) {
            if (i == 0) {
                results[i] = nums[i];
            } else {
                results[i] =results[i-1]+ nums[i];
            }
        }
        return results;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

}